What Is the Radius of the n = 6 Bohr Orbit in O7+?

[rayfieca]

Homework Statement

Calculate the radius of the n = 6 Bohr orbit in O7+(oxygen with 7 of its 8 electrons removed).
A) 190 pm B) 167 pm C) 238 pm D) 214 pm

Homework Equations

I believe that the relevant equation is
r(sub n)=(n^2)*a(sub b) where a(sub b)= Bohr's radius= 5.29*10^-11 m

The Attempt at a Solution

I calculated it for what I believed to be n=6, but I got an incorrect answer. The correct answer is (C) 238 pm, but I do not understand why.


Any help is greatly appreciated!

 

[hage567]

Your formula is not quite complete. You need to take into account the charge in the nucleus (so the number of protons).

 

[Redbelly98]

Welcome to PF

Your equation is for hydrogen, with a charge of +1 on the nucleus.
Since oxygen has a charge of Z=8 for the nucleus, that equation should be different, containing Z somehow.

Does your textbook discuss "hydrogen-like" or "hydrogenic" ions?

EDIT: ah, I should know better than to wait a 1/2 hour and then respond without refreshing the page.

 

[rayfieca]

Unfortunately,
I cannot find an equation that takes into account (Z). Anyone have the equation handy?
Thanks for all your help!

 

[hage567]

Look through this: http://hyperphysics.phy-astr.gsu.edu/hbase/bohr.html#c4

 

[rayfieca]

Thank you very much!